Cubic Twins

Using Cardano's method for cubic equation, we substitute:

$\begin{cases} x = y+1 \\ x= z+1 \end{cases} \Rightarrow \begin{cases} f(y) = y^3+2y - 14 \\ g(z) = z^3+2y +14 \end{cases}$

Since $a$ and $b$ are real roots of $f(x)$ and $g(x)$ respectively, when:

$\begin{cases} a = y+1 \\ b = z+1 \end{cases} \Rightarrow \begin{cases} y^3+2y = 14 \\ z^3+2y =-14 \end{cases}$

Adding the last two equations together:

$y^3+z^3+2(y+z) = 0\quad \Rightarrow (y+z)(y^2+z^2-yz+2) = 0$

We note that $y^2+z^2$ is always positive and by AM-GM equality, we know that $y^2+z^2 \ge 2 yz\quad \Rightarrow y^2+z^2-yz+2 > 0$ .

This means that $y+z = 0 \quad \Rightarrow a-1+b-1=0\quad \Rightarrow a+b = \boxed{2}$

Start with: $a^3-3a^2+5a-17=0$ $b^3-3b^2+5b+11=0$ Add the equations above to get: $a^3+b^3-3(a^2+b^2)+5(a+b)-6=0$ $\Rightarrow a^3+3ab(a+b)+b^3 -3(a^2+2ab+b^2)+5(a+b)-6-(3ab(a+b)+6ab)=0$ $\Rightarrow (a+b)^3-3(a+b)^2+5(a+b)-6-3ab(a+b-2)=0$ $\Rightarrow (a+b-2)((a+b)^2-(a+b)+3-3ab)=0$ But note that $(a+b)^2-(a+b)+3-3ab=(a+b)^2-(a+b)+3-3\Big(\Big(\frac{a+b}{2}\Big)^2-\Big(\frac{a-b}{2}\Big)^2\Big)$ $=\frac{1}{4}(a+b)^2-(a+b)+3+\frac{3}{4}(a-b)^2=\frac{1}{4}(a+b-2)^2+2+\frac{3}{4}(a-b)^2.$ If we assume that $a$ and $b$ are real, then this expression is strictly positive. Therefore, if $a$ and $b$ are the real roots of the equation, we must have $a+b-2=0$ or $a+b=2$ . $$ As a bonus, I will show that the complex roots taken in the right pairs also add up to $2$ . First of all, we already know by the graph that each cubic has only one real solution. But one can always test by taking the discriminant: $b^2c^2-4ac^3-3b^3d-27a^2d^2+18abcd=-5324,-5324$ . Since it is negative for both cubics, each has exactly one real root. Next, assume $x$ and $y$ are real numbers such that $y\neq 0$ , and substitute $x+iy$ into each cubic to set them to zero. I will write them both as one equation for convenience. $(x+iy)^3-3(x+iy)^2+5(x+iy)+[-17,11]=0$ $\Rightarrow x^3+3x^2iy-3xy^2-y^3-3x^2-6ixy+3y^2+5x+5iy+[-17, 11]=0$ Both the real and imaginary parts must be zero, so this gives two equations. $x^3-3xy^2-3x^2+3y^2+5x+[-17,11]=0$ $y^2=3x^2-6x+5$ Then we can substitute $y^2$ from the second equation into the first to obtain: $4x^3-12x^2+14x+[1,-13]=0$ $y^2=3x^2-6x+5$ The discriminant of the first equation(s) is equal to -21296 in both cases. So they have only one real root. (Also note that $y^2=3x^2-6x+5>0$ for any real $x$ . This ensures that the root in question exists as an imaginary root.) The next step is to, similar to the beginning of this solution, consider the pair of equations: $4u^3-12u^2+14u+1=0$ $4v^3-12v^2+14v-13=0$ Add them and then do some manipulations similar to before. $4(u^3+v^3)-12(u^2+v^2)+14(u+v)-12=0$ $\Rightarrow 2(u^3+v^3)-6(u^2+v^2)+7(u+v)-6=0$ $\Rightarrow 2(u+v)^3-6uv(u+v)-6(u+v)^2+12uv+7(u+v)-6=0$ $\Rightarrow 2(u+v)^3-6(u+v)^2+7(u+v)-6-6uv(u+v-2)=0$ $\Rightarrow (u+v-2)(2(u+v)^2-2(u+v)+3-6uv)=0$ Since $2(u+v)^2-2(u+v)+3-6uv=2(u+v)^2-2(u+v)+3-6\Big(\Big(\frac{u+v}{2}\Big)^2-\Big(\frac{u-v}{2}\Big)^2\Big)$ $=\frac{1}{2}(u+v)^2-2(u+v)+3+\frac{3}{2}(u-v)^2=\frac{1}{2}(u+v-2)^2+1+\frac{3}{2}(u-v)^2>0$ , we must have $u+v=2$ . This shows that the real parts of the imaginary roots of the two equations add to $2$ . Now I will show that the imaginary parts are equal in magnitude. First note that all the roots (and, in particular, the real roots) of the pair of equations $4u^3-12u^2+14u+1=0$ , $4v^3-12v^2+14v-13=0$ are distinct (this can be seen by substituting a root of one equation into the other and getting a contradiction). In other words $u-v\neq 0$ . Therefore, $u+v=2 \Leftrightarrow 3(u+v)-6=0 \Leftrightarrow 3(u+v)(u-v)-6(u-v)=0 \Leftrightarrow 3u^2-3v^2-6u+6v=0$ $\Leftrightarrow 3u^2-6u+5=3v^2-6v+5$ . Each side of the equation is precisely the square of the imaginary part for each solution. Therefore, they are equal in magnitude. To sum up, the roots can be written in the form { $a, u+iy, u-iy$ } and { $b, v+iy, v-iy$ }, where $a+b=2$ and $u+v=2$ .

What a great problem!

You don't need Cardano's method. See below.

$b^3-3b^2+5b+11=0$

From the above equations we can conclude that a,b are the roots of ,

An easy check confirms that 1 is also the root. Hence